Consider the polar curve $r=4\sin(5\theta)$. What is the equation of the tangent line to the curve $r$ at $\theta=\dfrac{\pi}{3}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $y-3=-\dfrac{\sqrt{3}}{2}(x+\sqrt{3})$ (Choice B) B $y+3=-\dfrac{\sqrt{3}}{2}(x+\sqrt{3})$ (Choice C) C $y+3=\dfrac{\sqrt{3}}{2}(x+\sqrt{3})$ (Choice D) D $y-3=\dfrac{\sqrt{3}}{2}(x+\sqrt{3})$
Explanation: The slope of the tangent line at a point is equal to $\dfrac{dy}{dx}$ at that point. In the case of polar curves, we can use the relationship: $\dfrac{dy}{dx}=\dfrac{\left( \dfrac{dy}{d\theta} \right)}{\left( \dfrac{dx}{d\theta} \right)}$ Then we can use the point-slope form to complete the equation for the tangent line through $\theta=\dfrac{\pi}{3}$. For a polar curve, $x={r}\cos(\theta)$ and $y={r}\sin(\theta)$. Therefore, in our problem we have: $\begin{aligned} x&={4\sin(5\theta)}\cos(\theta) \\\\ y&={4\sin(5\theta)} \sin(\theta) \end{aligned}$ Let's find $\dfrac{dy}{dx}$. $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{\left( \dfrac{dy}{d\theta} \right)}{\left( \dfrac{dx}{d\theta} \right)} \\\\ &=\dfrac{20\cos(5\theta)\sin(\theta)+4\sin(5\theta)\cos(\theta)}{20\cos(5\theta)\cos(\theta)-4\sin(5\theta)\sin(\theta)} \\\\ &=\dfrac{5\cos(5\theta)\sin(\theta)+\sin(5\theta)\cos(\theta)}{5\cos(5\theta)\cos(\theta)-\sin(5\theta)\sin(\theta)} \end{aligned}$ Evaluating $\dfrac{dy}{dx}$ at ${\theta = \dfrac{\pi}{3}}$ gives us the slope of our tangent line. $\begin{aligned} {\left. \dfrac{dy}{dx}\right| _{\theta=\tfrac{\pi}{3}}} &=\dfrac{5\cos\left(5\left({\dfrac{\pi}{3}}\right)\right)\sin\left({\dfrac{\pi}{3}}\right)+\sin\left(5\left({\dfrac{\pi}{3}}\right)\right)\cos\left({\dfrac{\pi}{3}}\right)}{5\cos\left(5\left({\dfrac{\pi}{3}}\right)\right)\cos\left({\dfrac{\pi}{3}}\right)-\sin\left(5\left({\dfrac{\pi}{3}}\right)\right)\sin\left({\dfrac{\pi}{3}}\right)} \\\\&=\dfrac{5\cos\left({\dfrac{5\pi}{3}}\right)\sin\left({\dfrac{\pi}{3}}\right)+\sin\left({\dfrac{5\pi}{3}}\right)\cos\left({\dfrac{\pi}{3}}\right)}{5\cos\left({\dfrac{5\pi}{3}}\right)\cos\left({\dfrac{\pi}{3}}\right)-\sin\left({\dfrac{5\pi}{3}}\right)\sin\left({\dfrac{\pi}{3}}\right)} \\\\ &=\dfrac{5\left(\dfrac{1}{2}\right)\left(\dfrac{\sqrt{3}}{2}\right)+\left(-\dfrac{\sqrt{3}}{2}\right)\left(\dfrac{1}{2}\right)}{5\left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)-\left(-\dfrac{\sqrt{3}}{2}\right)\left(\dfrac{\sqrt{3}}{2}\right)} \\\\ &=\dfrac{5\sqrt{3}-\sqrt{3}}{5+3} \\\\&={\dfrac{\sqrt{3}}{2}} \end{aligned}$ We now find $x$ and $y$ at the point $\theta=\dfrac{\pi}{3}$. $\begin{aligned} {x\left(\dfrac{\pi}{3}\right)}&=4\sin\left(\dfrac{5\pi}{3}\right)\cos\left(\dfrac{\pi}{3}\right) \\\\ &=4\left(-\dfrac{\sqrt{3}}{2}\right)\left(\dfrac{1}{2}\right) \\\\ &={-\sqrt{3}} \\\\ \\\\ y\left(\dfrac{\pi}{3}\right)}&=4\sin\left(\dfrac{5\pi}{3}\right)\sin\left(\dfrac{\pi}{3}\right) \\\\ &=4\left(\dfrac{\sqrt{3}}{2}\right)\left(\dfrac{\sqrt{3}}{2} \right) \\\\ &=-3} \end{aligned}$ Therefore the equation of our tangent line is : $\begin{aligned} y-(-3})&={\dfrac{\sqrt{3}}{2}}(x-({\sqrt{3}})) \\\\ y+3&=\dfrac{\sqrt{3}}{2}(x+\sqrt{3}) \end{aligned}$ The graph of the tangent is shown.